#### Question

In cyclic quadrilateral ABCD; AD = BC, ∠BAC = 30° and ∠CBD = 70°; find:

(i) ∠BCD (ii) ∠BCA (iii) ∠ABC (iv) ∠ADC

#### Solution

ABCD is a cyclic quadrilateral and AD = BC

∠BAC = 30° , ∠CBD = 70°

∠DAC = ∠CBD [Angles in the same segment]

⇒ ∠DAC = 70° [ ∵ CBD = 70°]

⇒ ∠BAD = ∠BAC + ∠DAC = 30° + 70° = 100°

Since the sum of opposite angles of cyclic quadrilateral is supplementary

∠BAD + ∠BCD =180°

⇒ 100°+ ∠BCD = 180° [From (1)]

⇒ ∠BCD = 180° -100° = 80°

Since , AD = BC ,∠ACD = ∠BDC [Equal chords subtends equal angles]

But ∠ACB = ∠ADB [angles in the same segment]

∴ ∠ACD + ∠ACB = ∠BDC + ∠ADB

⇒ ∠BCD = ∠ADC = 80°

But in ∆BCD,

∠CBD + ∠BCD + ∠BDC = 180 [angles oaf a triangle]

⇒ 70° + 80° + ∠BDC = 180°

⇒ 150° + ∠BDC = 180°

∴ ∠BDC = 180° -150° = 30°

⇒ ∠ACD = 30° [∠ACD = ∠BDC ]

∴ ∠BCA = ∠BCD - ∠ACD = 80° - 30° = 50°

Since the sum of opposite angles of cyclic quadrilateral is supplementary

∠ADC + ∠ABC =180°

⇒ 80° + ABC = 180°

⇒ ∠ABC =180° - 80° = 100°